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$\cos^{-1}\left(\cos\tfrac{7\pi}{6}\right)$ equals

A$\tfrac{7\pi}{6}$
B$-\tfrac{\pi}{6}$
C$\tfrac{\pi}{6}$
D$\tfrac{5\pi}{6}$
Answer & Solution
Correct answer: D. $\tfrac{5\pi}{6}$
$\cos^{-1}$ has range $[0, \pi]$, but $\tfrac{7\pi}{6} > \pi$. **Reflect about the x-axis:** $\cos\tfrac{7\pi}{6} = \cos\left(2\pi - \tfrac{7\pi}{6}\right) = \cos\tfrac{5\pi}{6} = -\tfrac{\sqrt{3}}{2}$, and $\tfrac{5\pi}{6} \in [0,\pi]$. **So the inverse returns $\tfrac{5\pi}{6}$.**
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